Flipping a matrix? Real quick question:
I was wondering, how would one denote mathemathically the flipping of a matrix, horizontally or vertically, around its own axis?
 A: It seems there is a general agreement that no such notation already exists. As I suggested in a comment you could use $A^H$ and $A^V$. If I was writing in a document in French, I would use ${}^h\!A$ and ${}^v\!A$. 
A: Let $P$ be the matrix with $1$ on the antidiagonal and $0$ otherwise, that is
$$P=\begin{pmatrix}0&0&\ldots& 0& 1\\0&0&\ldots & 1 & 0\\\vdots&\vdots &{}_.\cdot{}^\cdot&\vdots&\vdots\\
0&1&\ldots& 0& 0\\1&0&\ldots& 0& 0\\ \end{pmatrix}.$$ Then $AP$ and $PA$ (and $PAP$) are flipped versions of $A$.
A: Old question, but it is the first hit in search results for notation of flipping a matrix - so let's leave an answer.
Let $A \in \mathbb{R}^{m \times n}$. If you use the notation that matrix $A = A_{ij}$ where $i \in \{1,\ldots,m\}$ represents the row and $j \in \{1,\ldots,n\}$ represents the column. 
The following $7$ flips of $A$ come out of changing the indicies:


*

*$A^\top = A_{ji}$


*

*Transpose A by changing the order of the indices 


*$A_{i(n+1-j)}$.


*

*Flip $A$ horizontally 

*This interchanges the columns so that the first column of $A$ is the last column of $A_{i(n+1-j)}$, and the last column of $A$ is the first column of $A_{i(n+1-j)}$. 


*$A_{(m+1-i)j}$


*

*Flip $A$ vertically

*This interchanges the rows so that the first row of $A$ is the last row of $A_{(m+1-i)j}$, and the last row of $A$ is the first row of $A_{(m+1-i)j}$.


*$A_{(m+1-i)(n+1-j)}$


*

*Flip $A$ horizontally then vertically

*Same as flip vertical then horizontal


*$A_{(n+1-j)i}$


*

*Flip $A^\top$ vertically


*$A_{j(m+1-i)}$


*

*Flip $A^\top$ horizontally 


*$A_{(n+1-j)(m+1-i)}$


*

*Flip $A^\top$ vertically then horizontally


